Uniform continuity and normality of metric spaces in $\mathbf{ZF}$

Kyriakos Keremedis¹

¹ Department of Mathematics University of the Aegean Karlovassi, Samos 83200, Greece

Bulletin of the Polish Academy of Sciences. Mathematics, Tome 65 (2017) no. 2, pp. 113-124

Résumé

Let $\mathbf{X}=(X,d)$ and $\mathbf{Y}=(Y,\rho )$ be two metric spaces. (a) We show in $\mathbf{ZF}$ that: (i) If $\mathbf{X}$ is separable and $f:\mathbf{X}\rightarrow \mathbf{Y}$ is a continuous function then $f$ is uniformly continuous iff for any $ A,B\subseteq X$ with $d(A,B)=0$, $\rho (f(A),f(B))=0$. But it is relatively consistent with $\mathbf{ZF}$ that there exist metric spaces $\mathbf{X}$, $ \mathbf{Y}$ and a continuous, non-uniformly continuous function $f:\mathbf{X} \rightarrow \mathbf{Y}$ such that for any $A,B\subseteq X$ with $d(A,B)=0$, $\rho (f(A),f(B))=0$. (ii) If $S$ is a dense subset of $\mathbf{X}$, $\mathbf{Y}$ is Cantor complete and $f:\mathbf{S}\rightarrow \mathbf{Y}$ a uniformly continuous function, then there is a unique uniformly continuous function $F:\mathbf{X} \rightarrow \mathbf{Y}$ extending $f$. But it is relatively consistent with $\mathbf{ZF}$ that there exist a metric space $\mathbf{X}$, a complete metric space $\mathbf{Y}$, a dense subset $S$ of $\mathbf{X}$ and a uniformly continuous function $f:\mathbf{S} \rightarrow \mathbf{Y}$ that does not extend to a uniformly continuous function on $\mathbf{X}$. (iii) $\mathbf{X}$ is complete iff for any Cauchy sequences $ (x_{n})_{n\in \mathbb{N}}$ and $(y_{n})_{n\in \mathbb{N}}$ in $\mathbf{X}$, if $\overline{ \{x_{n}:n\in \mathbb{N}\}}\cap \overline{\{y_{n}:n\in \mathbb{N}\}} =\emptyset $ then $d(\{x_{n}:n\in \mathbb{N}\},\{y_{n}:n\in \mathbb{N} \}) \gt 0 $. (b) We show in $\mathbf{ZF}$+$\mathbf{CAC}$ that if $f:\mathbf{X} \rightarrow \mathbf{Y}$ is a continuous function, then $f$ is uniformly continuous iff for any $A,B\subseteq X$ with $d(A,B)=0$, $\rho (f(A),f(B))=0$.

Détail

DOI : 10.4064/ba8122-10-2017

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